Question: Simplify and expand the following expression: $ \dfrac{2}{a + 9}+ \dfrac{2}{3a - 9}- \dfrac{1}{a^2 + 6a - 27} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{2}{3a - 9} = \dfrac{2}{3(a - 3)}$ We can factor the quadratic in the third term: $ \dfrac{1}{a^2 + 6a - 27} = \dfrac{1}{(a + 9)(a - 3)}$ Now we have: $ \dfrac{2}{a + 9}+ \dfrac{2}{3(a - 3)}- \dfrac{1}{(a + 9)(a - 3)} $ The least common multiple of the denominators is: $ (a + 9)(a - 3)$ In order to get the first term over $(a + 9)(a - 3)$ , multiply by $\dfrac{3(a - 3)}{3(a - 3)}$ $ \dfrac{2}{a + 9} \times \dfrac{3(a - 3)}{3(a - 3)} = \dfrac{6(a - 3)}{(a + 9)(a - 3)} $ In order to get the second term over $(a + 9)(a - 3)$ , multiply by $\dfrac{a + 9}{a + 9}$ $ \dfrac{2}{3(a - 3)} \times \dfrac{a + 9}{a + 9} = \dfrac{2(a + 9)}{(a + 9)(a - 3)} $ In order to get the third term over $(a + 9)(a - 3)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{1}{(a + 9)(a - 3)} \times \dfrac{3}{3} = \dfrac{3}{(a + 9)(a - 3)} $ Now we have: $ \dfrac{6(a - 3)}{(a + 9)(a - 3)} + \dfrac{2(a + 9)}{(a + 9)(a - 3)} - \dfrac{3}{(a + 9)(a - 3)} $ $ = \dfrac{ 6(a - 3) + 2(a + 9) - 3} {(a + 9)(a - 3)} $ Expand: $ = \dfrac{6a - 18 + 2a + 18 - 3}{3a^2 + 18a - 81} $ $ = \dfrac{8a - 3}{3a^2 + 18a - 81}$